Respuesta :
Answer:
a) E_net = 0
b) E_net = 2663.7 N/C
c) E_net = 525.915 N/C
d) E_net = 1380.864 N/C
Explanation:
Given:
- The positions of charges are = ( 0.15 , 0 ) & (-0.15 , 0 )
- Both charges have equal magnitude Q = +6*10^-19 C
Find:
Find the magnitude of the electric field at:
(a) the origin;
(b) x = 0.300 m, y = 0;
(c) x = 0.150 m, y = -0.400 m;
(d) x = 0, y = 0.200 m.
Solution:
- The Electric field generated by an charged particle Q at a distance r is given by:
E = k*Q / r^2
- Where, k is the coulomb's constant = 8.99 * 10^9
Part a) @ ( 0 , 0 )
- First we see that the charges +Q_1 and +Q_2 produce an electric field equal but opposite in nature because the distance r from origin and magnitude of charge is same. So the sum of Electric fields:
E_1 + E_3 = 0
Part b) @ ( 0.3 , 0 )
- The Electric field due to charge Q_1 is expressed by:
E_1 = k*Q_1 / r_1^2
Where the distance r from first charge to ( 0.3 , 0 ) is:
r_1 = 0.3 - 0.15 = 0.15 m
- The Electric field due to charge Q_2 is expressed by:
E_2 = k*Q_2 / r_2^2
Where the distance r from second charge to ( 0.3 , 0 ) is:
r_2 = 0.3 - (-0.15) = 0.45 m
- The net Electric Field at point ( 0.3 , 0 ) is:
E_net = E_1 + E_2
E_net = k*Q* ( ( 1 / r_1^2) + ( 1 / r_2^2) )
Plug values in:
E_net = (8.99*10^9)*6*10^-9* ( ( 1 / 0.15^2) + ( 1 / 0.45^2)))
E_net = 2663.7 N/C
Part c) @ ( 0.15 , -0.4 )
- The Electric field due to charge Q_1 is expressed by:
E_1 = k*Q_1 / r_1^2
Where the distance r from first charge to ( 0.15 , -0.4 ) is:
r_1 = 0.4 m
- The Electric field due to charge Q_2 is expressed by:
E_2 = k*Q_2 / r_2^2
Where the distance r from second charge to ( 0.15 , -0.4 ) is:
r_2 = sqrt (0.3^2 + 0.4^2) = 0.5 m
- The net Electric Field at point ( 0.15 , -0.4 ) is:
E_vertical = E_1 + E_2*sin(a)
E_horizontal = E_2*cos(a)
Where, a is the angle between x -axis and point ( 0.15 , -0.4 ):
cos(a) = 0.3 / r_2 = 0.3/0.5 = 3/5
sin(a) = 0.4 / r_2 = 0.4/0.5 = 4/5
Hence,
E_vertical = k*Q* ( ( 1 / r_1^2) + ( sin(a)/ r_2^2)
E_v = (8.99*10^9)*6*10^-9* ( ( 1 / 0.4^2) + ( 0.8 / 0.5^2 )
E_v = 509.733 N/C
And,
E_Horizontal = k*Q* ( ( 1 / r_1^2) + ( sin(a)/ r_2^2)
E_h = (8.99*10^9)*6*10^-9*( 0.6 / 0.5^2 )
E_h = 129.456 N/C
Hence,
E_net = sqrt ( E_vertical^2 + E_horizontal^2)
E_net = sqrt ( 509.733^2 + 129.456^2)
E_net = 525.915 N/C
Part d) @ ( 0 , 0.2 )
- The Electric field due to charge Q_1 is expressed by:
E_1 = k*Q_1 / r_1^2
Where the distance r from first charge to ( 0 , 0.2 ) is:
r_1 = sqrt (0.15^2 + 0.2^2) = 0.25 m
- The Electric field due to charge Q_2 is expressed by:
E_2 = k*Q_2 / r_2^2
Where the distance r from second charge to ( 0 , 0.2 ) is:
r_2 = sqrt (0.15^2 + 0.2^2) = 0.25 m
- The net Electric Field at point ( 0 , 0.2 ) is:
E_vertical = E_1*sin(a) + E_2*sin(a)
E_vertical = 2*E*sin(a)
E_horizontal = 0 ... Equal but opposite magnitude
Where, a is the angle between x -axis and point ( 0 , 0.2 ):
sin(a) = 0.2 / r_2 = 0.2/0.25 = 4/5
Hence,
E_vertical = 2*k*Q*sin(a)*( 1 / r_1^2)
E_v = 2*0.8*(8.99*10^9)*6*10^-9* ( 1 / 0.25^2)
E_v = 1380.864 N/C
Hence,
E_net = E_vertical
E_net = 1380.864 N/C
